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Conservation of Energy. Fountain Question

  • Thread starter mfianist
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  • #1
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Water shoots up from a fountainat a speed of v0 at a rate of dm/dt, and hits a garbage can of mass M suspended in the air. What is the maximum height the fountain can lift the garbage can?

First i approached this problem with K.E = P.E

i assumed that m at time t = (dm/dt)t

so i put {M+(dm/dt)t}gh = 1/2 {(dm/dt)tv^2}

then i solved the equation for h.

i am not so sure if my approach is right...


Is there a better of solving this problem?
 

Answers and Replies

  • #2
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Interesting problem. I was looking at this as an impulse problem. Recall the formula impulse P = Ft? So, assuming constant F, this gives us dP = Fdt for very small impulse bits. Rearrange and you have F = dP/dt, or, force equals the time derivative of momentum. Ah, momentum. So, what is the momentum of an object thrown upward as a function of its height? P=mv, so what is v as a function of h and some initial velocity v0? Find this P as a function of h, then find dP/dt keeping in mind dP = mdv + dmv and v is constant for a particular h. So, the garbage can has mass M, weighs Mg, it's force downward is its weight, where does the equivalent force upward come from?

This would probably be my approach, anyone see if I missed something obvious? I'm assuming inelastic collisions with the water particles, which seems reasonable.
 

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